Question

two charges Q1 is equals to + 48 microcoulomb and Q2 is equals to minus 54 microcoulombs are kept 0.6 millimetre apart find the force acting between them. give its nature​

Answer 1

Answer:

F = -6.48×10^7 N

Explanation:

Given, Q1 = +48 × 10^-6 C

Q2 = -54 × 10^-6 C

d = 6 × 10^-4 m

$f = \frac{1}{4\pi \: e} \frac{q1 \times q2}{ {d }^{2} }$

F = 9×10^9 ×(48×10^-6)×(-54 × 10^-6 C)/36×10^-8

F = 9×10^-3 x(48×-54)/36×10^-8

F = -(648) × 10^5

F = -6.48×10^7 N

Since the charges are opposite,the force is attractive.